Homework Help: Yet Another Delta-Epsilon

1. Jul 22, 2010

zooxanthellae

1. The problem statement, all variables and given/known data

Prove that lim (1/x) x --> 0 does not exist, i.e., show that lim (1/x) x --> 0 = l is false for every number l.

2. Relevant equations
0 < |x-a| < d
|f(x) - l| < E

3. The attempt at a solution

The strange thing is, the first time through I got the same solution as Spivak, but looking over it again the logic seems downright wrong. Here's the solution, verbatim:

Where is he getting the bold portion from?

I write |1/x - l| < E
|1/x| - |l| < |1/x - l| < E
|1/x| - |l| < E
|1/x| < E + |l|

How is he getting greater-than?

2. Jul 22, 2010

Office_Shredder

Staff Emeritus

Assuming this is true, then you're golden. But it's not going to be true for really small x, that's the point. So we see how he's doing a proof by contradiction: If the limit exists, you can't possibly get a greater than sign, but he gets a greater than sign. The way he gets it is by saying when x < min(d, 1/(|l| + E)), you get 1/x > |l| + E

3. Jul 22, 2010

vela

Staff Emeritus
You're misreading what he wrote. He is merely asserting that there is an x such that 0<x<d and 1/x>|l|+E. Then he tells you how to find such an x.
If the inequality you started with is true, then your conclusion is also true. If you go one step further and take the reciprocal of both sides, you get

$$|x| > \frac{1}{|l|+\epsilon}$$

Spivak is saying if you choose an x such that this last inequality is false, the one you started with will also be false.

4. Jul 22, 2010

zooxanthellae

So, he's essentially saying:

|1/x| < E + |l|

and then pointing out that, if d = 1/(|l| + E), this leads to a contradiction of that inequality, and therefore the whole thing violates the definition of a limit?

Oy! It's been almost a week and I'm still having trouble with these things.

5. Jul 22, 2010

Office_Shredder

Staff Emeritus
He isn't saying that because it's not going to be true. But it's what you should keep in the back of your mind... IF l was the limit, you would get this inequality. But in fact we get an opposite inequality for small x. Therefore, l is not the limit. Since l was arbitrary, there is no limit

6. Jul 22, 2010

vela

Staff Emeritus
No, he's stating if 0<|x|<min{δ, 1/(|L|+ε)}, then 0<|x|<δ and |1/x|>|L|+ε. The δ appears only because the definition of the limit requires that |x|<δ. So long as |x| is small enough, you can satisfy both conditions. The point is, regardless of what δ equals, you can always find an |x| between 0 and δ such that |1/x|>|L|+ε.

When you try to come up with a delta-epsilon proof, typically you work backwards. You assume the epsilon condition holds and find some condition on x from which you can choose an appropriate δ. In this case, you can't find such a δ because the inequality turns out backwards.

7. Jul 22, 2010

Buri

I have those kinds of days too :) lol

8. Jul 23, 2010

zooxanthellae

Let me try again: because there are values for delta (e.g. <1/(|L| + E)) that violate the inequality that would result if the limit were true, then the limit is not true by contradiction?

9. Jul 23, 2010

vela

Staff Emeritus
No, that's not right. Even when a limit exists, you can usually find a δ for which the ε-inequality won't hold: just use a big δ. The limit doesn't exist because there are no values of δ for which the ε-inequality holds.

10. Jul 23, 2010

zooxanthellae

Yeah, I think I figured it out a few hours ago (I was doing another problem and realized what I said here was wrong). It's because we've found a d within which no x will work (all x results in an inconsistency, no matter how small; in fact, the smaller the x the worse the inconsistency becomes)?

11. Jul 23, 2010

vela

Staff Emeritus
Almost. You don't need to find a δ such that all values of x within the interval will cause the ε inequality to fail, though in this case you can. You only need one value of x for which inequality fails.